SOLUTIONS TO EXERCISES

CHEMISTRY

THE CENTRAL SCIENCE

14 ^TH EDITION

SOLUTIONS TO EXERCISES

Roxy Wilson

University of Illinois, Urbana-Champaign

CHEMISTRY

THE CENTRAL SCIENCE

14 ^TH EDITION

BROWN | L E MAY

BURSTEN | MURPHY

WOODWARD | STOLTZFUS

 

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1 17 ISBN-10: 0-13-455224-5 ISBN-13: 978-0-13-455224-8

Contents

Chapter 1 Introduction: Matter, Energy, and Measurement

Chapter 2 Atoms, Molecules, and Ions

Chapter 3 Chemical Reactions and Reaction Stoichiometry

Chapter 4 Reactions in Aqueous Solution

Chapter 5 Thermochemistry

Chapter 6 Electronic Structure of Atoms

Chapter 7 Periodic Properties of the Elements

Chapter 8 Basic Concepts of Chemical Bonding

Chapter 9 Molecular Geometry and Bonding Theories

Chapter 10 Gases

Chapter 11 Liquids and Intermolecular Forces

Chapter 12 Solids and Modern Materials

Chapter 13 Properties of Solutions

Chapter 14 Chemical Kinetics

Chapter 15 Chemical Equilibrium

Chapter 16 Acid–Base Equilibria

Chapter 17 Additional Aspects of Aqueous Equilibria

Chapter 18 Chemistry of the Environment

Chapter 19 Chemical Thermodynamics

Chapter 20 Electrochemistry

Chapter 21 Nuclear Chemistry

Chapter 22 Chemistry of the Nonmetals

Chapter 23 Transition Metals and Coordination Chemistry

Chapter 24 The Chemistry of Life: Organic and Biological Chemistry

Introduction

Chemistry: The Central Science, 14th edition , contains over 2700 end-of-chapter exercises. Considerable attention has been given to these exercises because one of the best ways for students to master chemistry is by solving problems. Grouping the exercises according to subject matter is intended to aid the student in selecting and recognizing particular types of problems. Within each subject matter group, similar problems are arranged in pairs. This provides the student with an opportunity to reinforce a particular kind of problem. There are also a substantial number of general exercises in each chapter to supplement those grouped by topic. Integrative exercises, which require students to integrate concepts from several chapters, are a continuing feature of the 14th edition. Answers to the odd-numbered topical exercises plus selected general and integrative exercises, about 1150 in all, are provided in the textbook. These appendix answers help to make the textbook a useful self-contained vehicle for learning.

This manual, Solutions to Exercises in Chemistry: The Central Science, 14th edition , was written to enhance the end-of-chapter exercises by providing documented solutions. The manual assists the instructor by saving time spent generating solutions for assigned problem sets and aids the student by offering a convenient independent source to check their understanding of the material. Most solutions have been worked in the same detail as the in-chapter sample exercises to help guide students in their studies.

To reinforce the Analyze, Plan, Solve, Check problem-solving method used extensively in the text, this strategy has also been incorporated into the Solutions Manual. Solutions to most red topical exercises and selected Additional and Integrative exercises feature this four-step approach. We strongly encourage students to master this powerful and totally general method.

When using this manual, keep in mind that the numerical result of any calculation is influenced by the precision of the numbers used in the calculation. In this manual, for example, atomic masses and physical constants are typically expressed to four significant figures, or at least as precisely as the data given in the problem. If students use slightly different values to solve problems, their answers will differ slightly from those listed in the appendix of the text or this manual. This is a normal and a common occurrence when comparing results from different calculations or experiments.

Rounding methods are another source of differences between calculated values. In this manual, when a solution is given in steps, intermediate results will be rounded to the correct number of significant figures; however, unrounded numbers will be used in subsequent calculations. By following this scheme, calculators need not be cleared to re-enter rounded intermediate results in the middle of a calculation sequence. The final answer will appear with the correct number of significant figures. This may result in a small discrepancy in the last significant digit between student-calculated answers and those given in this manual. Variations due to rounding can occur in any analysis of numerical data.

The first step in checking your solution and resolving differences between your answer and the listed value is to look for similarities and differences in problem-solving methods. Ultimately, resolving the small numerical differences described above is less important than understanding the general method for solving a problem. The goal of this manual is to provide a reference for sound and consistent problem-solving methods in addition to accurate answers to text exercises.

Extraordinary efforts have been made to keep this manual as error-free as possible. All exercises were worked and proofread by at least three chemists to ensure clarity in methods and accuracy in mathematics. The ongoing work and advice of Dr. Richard Helmich, Ms. Rene Musto, and Dr. Christopher Musto continue to be invaluable to this project. In any written work as technically challenging as this manual, typos and errors inevitably creep in, despite our combined efforts. Please help us find and eliminate them. We hope that both instructors and students will find this manual accurate, helpful, and instructive.

Roxy B. Wilson, Ph.D. 1829 Maynard Dr. Champaign, IL 61822 rbwilson@illinois.edu

 

1 Introduction: Matter, Energy, and Measurement

Visualizing Concepts

1.1 Pure elements contain only one kind of atom. Atoms can be present singly or as tightly bound groups called molecules. Compounds contain two or more kinds of atoms bound tightly into molecules. Mixtures contain more than one kind of atom and/or molecule, not bound into discrete particles.

1. pure element: i
2. mixture of elements: v, vi
3. pure compound: iv
4. mixture of an element and a compound: ii, iii

1.2   After a physical change , the identities of the substances involved are the same as their identities before the change. That is, molecules retain their original composition. During a chemical change , at least one new substance is produced; rearrangement of atoms into new molecules occurs.

Diagram (b) represents a chemical change because the molecules after the change are different than the molecules before the change.

1.3

1. Brass is composed of two different kinds of atoms, so it is a mixture. The mixture appears homogeneous under an optical microscope, so it is a homogeneous mixture.
2. Because brass is a homogeneous mixture, it is a solution. We usually think of solutions as liquids, but they can be solids, liquids, or gases.

1.4   Refer to Sample Exercise 1.5.

1. $\begin{array}{l}\text{density}=\frac{\text{mass}}{\text{volume}};\text{mass}=\text{density}\times \text{volume}\\ \text{mass(Al)}=\frac{2.70\text{g}}{{\text{cm}}^3}\times 196{\text{cm}}^3\times \frac{1\text{kg}}{1000\text{g}}=0.5292=0.529\text{kg}\\ \text{mass(Ag)}=\frac{10.49\text{g}}{{\text{cm}}^3}\times 196{\text{cm}}^3\times \frac{1\text{kg}}{1000\text{g}}=2.056=2.06\text{kg}\end{array}$
2. The amount of work done to lift each sphere to a height of 2.2 m can be calculated using Equation 1.1, w = F × d = m × g × d. Use mass in kg and distance in meters so the final unit for work is joules, J.

1. $\begin{array}{c}\text{w}(\text{Al})=0.5292\text{kg}\text{Al}\times \frac{9.8\text{m}}{{\text{s}}^2}\times 2.2\text{m}=11.41=\frac{11{\text{kg-m}}^2}{1{\text{s}}^2}=11\text{J}\\ \text{w(Ag)}=2.056\text{kg}\text{Ag}\times \frac{9.8\text{m}}{{\text{s}}^{\text{2}}}\times 2.2\text{m}=44.33=\frac{44{\text{kg-m}}^2}{{\text{s}}^2}=44\text{J}\end{array}$

   (The results have two significant figures because the height had only two significant figures.)

   Note: This is the first exercise where “intermediate rounding” occurs. In this manual, when a solution is given in steps, the intermediate result will be rounded to the correct number of significant figures. However, the unrounded number will be used in subsequent calculations. The final answer will appear with the correct number of significant figures. That is, calculators need not be cleared and new numbers entered in the middle of a calculation sequence. This may result in a small discrepancy in the last significant digit between student-calculated answers and those given in the manual. These variations occur in any analysis of numerical data.

   For example, in this exercise the mass of the Ag sphere, 2.056 × 10 ³ g, is rounded to 2.06 × 10 ³ g, but 2.056 × 10 ³ g is retained in the subsequent calculation of work, 44 J. In this case, the unrounded and rounded masses both lead to the same rounded value for work. In other exercises, the correctly rounded results of the two methods may not be identical.

1. Less work is done on the Al sphere, so its potential energy increases by a smaller amount than the potential energy of the Ag sphere.

1. No. The two spheres do not have the same kinetic energy when they touch the ground. Because the Ag sphere has greater mass, it gains more potential energy when it is lifted. At the point of impact, all potential energy has been converted to kinetic energy. Because of its greater potential energy, the Ag sphere has greater kinetic energy when it hits the floor.

1.5   Filtration. When brewing a cup of coffee, hot water contacts the coffee grounds and dissolves components of the coffee bean that are water-soluble. This creates a heterogeneous mixture of undissolved coffee bean solids and liquid coffee solution; this mixture is separated by filtration. Undissolved grounds remain on the filter paper and liquid coffee drips into the container below.

1.6

1. time
2. mass
3. temperature
4. area
5. length
6. area
7. temperature
8. density
9. volume

1.7   Density is the ratio of mass to volume. For a sphere, size is like volume; both are determined by the radius of the sphere.

1. For spheres of the same size or volume, the denominator of the density relationship is the same. The denser the sphere, the heavier it is. A list from lightest to heaviest is in order of increasing density and mass. The aluminum sphere (density = 2.70 g/cm ³ ) is lightest, then nickel (density = 8.90 g/cm ³ ), then silver (density = 10.49 g/cm ³ ).

1. For cubes of equal mass, the numerator of the density relationship is the same. The denser the sphere, the smaller its volume or size. A list from smallest to largest is in order of decreasing density. The platinum sphere (density = 21.45 g/cm ³ ) is smallest, then gold (density = 19.30 g/cm ³ ), then lead (density = 11.35 g/cm ³ ).

1.8   Results (bullet holes) that are close to each other are precise . Results that are close to the “true value” (the bull’s-eye) are accurate .

1. The results on target A are precise but not accurate. The individual shots landed close to each other, but not close to the bull’s-eye. The results on target B are both precise and accurate. The individual shots landed close together and in the center of the target. The results on target C are neither precise nor accurate. The individual shots are scattered around the target.
2. The precise grouping on A is high and to the right of center. To improve accuracy, the sighting mechanism on the gun should be adjusted down and slightly left from its current position.

   To improve the results on C, someone who can produce a precise grouping must shoot the student’s target rifle. Then, the position of the sighting mechanism can be adjusted to produce an accurate shot. After adjusting the sight, the student needs more practice to improve precision.

1.9

1. 7.5 cm. There are two significant figures in this measurement; the number of cm can be read precisely, but there is some estimating required to read tenths of a centimeter. Listing two significant figures is consistent with the convention that measured quantities are reported so that there is uncertainty in only the last digit.
2. The speed is 72 mi/hr (inner scale, two significant figures) or 115 km/hr (outer scale, three significant figures). Both scales are read with certainty in the “hundreds” and “tens” place, and some uncertainty in the “ones” place. The km/hr speed has one more significant figure because its magnitude is in the hundreds.

1.10

1. Volume = length × width × height. Because the operation is multiplication, the dimension with fewest significant figures (sig figs) determines the number of sig figs in the result. The dimension “2.5 cm” has 2 sig figs, so the volume is reported with 2 sig figs.
2. Density = mass/volume. Because the operation is division, again the datum with fewer significant figures determines the number of sig figs in the result. While mass, 104.72 g, has 5 sig figs, volume [from (a)] has 2 sig figs, so density is also reported to 2 sig figs.

1.11  Given: masses of six jelly beans, mass of full jar, mass of empty jar. Find: number of jelly beans in the jar. The total mass of jelly beans is the mass of the jar full minus the mass of the jar empty. The mass of an “average” jelly bean is the average of the six masses. Then, the number of jelly beans is the total mass of beans divided by the average mass of a single bean.

Total mass of beans = 2082 g – 653 g = 1429 g

Average mass of a bean = (3.15 + 3.12 + 2.98 + 3.14 + 3.02 + 3.09) / 6 = 3.08 g

Number of beans = 1429 g total / 3.08 g per bean = 463.96 = 464 beans

By applying the significant figure rules for addition and subtraction, the total mass of beans has 0 decimal places and thus 4 significant figures. The mass of an average bean has 2 decimal places and 3 sig figs. The number of beans then has 3 sig figs, by the rules for multiplication and division. This makes sense, because we expect an integer number of beans in the jar. (Note that using an unrounded average bean mass of 3.0833 g predicts the number of beans to be 463.46, which rounds to 463 beans. The difference in these two values shows uncertainty in the last significant figure of the number of beans, as we expect in an experimental result.)

1.12  Compounds are pure substances, they have constant composition and properties throughout. The agate stone cannot be a compound, because materials with different properties appear as irregular rings in the stone. Ellen is correct.

Classification and Properties of Matter (Sections 1.2 and 1.3)

1.13

1. heterogeneous mixture
2. homogeneous mixture (If there are undissolved particles, such as sand or decaying plants, the mixture is heterogeneous.)
3. pure substance
4. pure substance

1.14

1. homogeneous mixture
2. heterogeneous mixture (particles in liquid)
3. pure substance
4. heterogeneous mixture

1.15

1. S
2. Au
3. K
4. Cl
5. Cu
6. uranium
7. nickel
8. sodium
9. aluminum
10. silicon

1.16

1. C
2. N
3. Ti
4. Zn
5. Fe
6. phosphorus
7. calcium
8. helium
9. lead
10. silver

1.17  A(s) → B(s) + C(g)

Substances A and C are definitely compounds; B is probably a compound. When solid carbon is burned in excess oxygen gas, the two elements combine to form a gaseous compound, carbon dioxide. Clearly substance C is this compound. Because C is produced when A is heated in the absence of oxygen (from air), both the carbon and the oxygen in C must have been present in A originally. A is, therefore, a compound composed of two or more elements chemically combined. Without more information on the chemical or physical properties of B, we cannot determine absolutely whether it is an element or a compound. However, few if any elements exist as white solids, so B is probably also a compound.

1.18  Gold, Au, and “fool’s gold,” FeS ₂ , are similar in appearance and physical state, so these two properties are not useful for discriminating between the two substances. Density and melting point would both help determine if the nugget is gold. (For these two substances, density is easiest to measure and is definitive.)

1.19 Physical properties : silvery white (color); lustrous; melting point = 649 °C; boiling point = 1105 °C; density at 20 °C = 1.738 g/cm ³ ; pounded into sheets (malleable); drawn into wires (ductile); good conductor. Chemical properties : burns in air to give intense white light; reacts with Cl ₂ to produce brittle white solid.

1.20

1. Physical properties : melting point = 420 °C; hardness = 2.5 Mohs; density = 7.13 g/cm ³ at 25 °C. Chemical properties : granules react with dilute sulfuric acid to produce hydrogen gas; at elevated temperatures, reacts slowly with oxygen gas to produce ZnO.
2. From the photo, we see that zinc is a shiny dark gray solid at atmospheric conditions, (physical state, color). These are physical properties.

1.21

1. chemical
2. physical
3. physical
4. chemical
5. chemical

1.22

1. chemical
2. physical
3. physical (The production of H ₂ O is a chemical change, but its condensation is a physical change.)
4. physical (The production of soot is a chemical change, but its deposition is a physical change.)

1.23  Distillation. A solution of sugar and water is a homogeneous mixture that cannot be separated by filtration. Distillation takes advantage of the much lower boiling point of water.

1.24

1. Chemical. A substance with new chemical properties is produced when the two clear liquids are mixed. The product of physical mixing would be another colorless solution.
2. Of the three methods, filtration is the most convenient way to separate an insoluble solid from a liquid.

The Nature of Energy (Section 1.4)

1.25

1. Plan .  E _k = 1/2 mv ² ; m = 1200 kg; v = 18 m/s; 1 kg-m ² /s ² = 1 J

   Solve .  E _k = 1/2 × 1200 kg × (18) ² m ² /s ² = 1.944 × 10 ⁵ = 1.9 × 10 ⁵ J

2. $1\text{cal}=4.184\text{J;}1.944\times {10}^5\text{J}\times \frac{1\text{cal}}{4.184\text{J}}=4.646\times {10}^4\text{cal}=4.6\times {10}^4\text{cal}$
3. As the automobile brakes to a stop, its speed (and hence its kinetic energy) drops to 0. This “lost” kinetic energy is mostly converted to heat. (The heat shows up in the brake parts, tires, and road.) It is not converted to some form of potential energy.

1.26

1. Analyze .  Given: mass and speed of ball. Find: kinetic energy.

   Plan . Because 1 J = 1 kg-m ² /s ² convert oz to kg and mph to m/s to obtain E _k in J.

   Solve . $5.13\text{oz}\times \frac{1\text{lb}}{16\text{oz}}\times \frac{1\text{kg}}{2.205\text{lb}}=0.14541=0.145\text{kg}$

1. $\begin{array}{l}\frac{95.0\text{mi}}{1\text{hr}}\times \frac{1.6093\text{km}}{1\text{mi}}\times \frac{1000\text{m}}{1\text{km}}\times \frac{1\text{hr}}{60\text{min}}\times \frac{1\text{min}}{60\text{sec}}=42.468=42.5\text{m/s}\\ {\text{E}}_{\text{k}}=1/2{\text{mv}}^2=1/2\times 0.14541\text{kg}\times {\left(\frac{42.468\text{m}}{1\text{s}}\right)}^2=\frac{131{\text{kg-m}}^2}{1{\text{s}}^2}=131\text{J}\end{array}$

   Check .  1/2(0.15 × 1600) ≈ 1/2(160+80) ≈ 120 J

1. Kinetic energy is related to velocity squared (v ² ); if the speed of the ball decreases to 55.0 mph, the kinetic energy of the ball will decrease by a factor of (55.0/95.0) ² . (The conversion factors to m/s apply to both speeds and will cancel in the ratio.) The numerical multiplier is (55/95) ² = 0.335. The kinetic energy decreases by approximately a factor of 3.

1. As the ball hits the catcher’s glove, its speed (and hence its kinetic energy) drops to 0. Most of the energy is converted to heat, which is transferred to the glove and the hand inside. If the catcher’s arm (and shoulder) is considered a spring, some kineic energy is transferred to potential energy of the arm assembly, which recoils while catching the ball. [The event can also be considered an inelastic collision between the ball and the arm assembly, with the combined masses (arm+ ball) having the same kinetic energy and the arm doing work to decelerate the (arm + ball).]

1.27

1. Kinetic energy; the particles move apart.
2. Potential energy decreases. The greater the separation prior to release, the smaller the electrostatic repulsion and potential energy.

1.28

1. Increases. The magnitude of the electrostatic attractive potential energy gets smaller as the distance between the particles increases, but the sign is negative. The potential energy increases because it becomes less negative. From another perspective, energy must be supplied to separate particles that are attracted to each other. This energy leads to an increase in the potential energy of the particles.
2. Increases. Work must be done to pump water in opposition to the force of gravity. This work increases the potential energy of the water.
3. Increases. Energy must be supplied to break a chemical bond. The potential energy of the atoms increases.

1.29 Analyze/Plan. Use results from Solution 1.4 and energy relationships discussed in Section 1.4 to solve for kinetic energy and velocity.

Solve. From Solution 1.4 we have the work required to lift the spheres. This is equal to the potential energy of the spheres at 2.2 m. As a sphere hits the floor, all potential energy is changed to kinetic energy. The kinetic energy of the Al sphere is then 11 J.

E _k = 11.41 = 11 J; mass(Al) = 0.5292 = 0.529 kg; E _k = 1/2 mv ² ; v = (2 E _k /m) ^1/2 ; $\text{v(Al)}={\left(2\times \frac{\text{11}\text{.41}{\text{kg-m}}^{\text{2}}}{{\text{s}}^{\text{2}}}\times \frac{\text{1}}{\text{0}\text{.5292}\text{kg}}\right)}^{1/2}=6.567=6.6\text{m/s}$

1.30  E _k = 44.33 = 44 J; mass(Ag) = 2.056 kg = 2.06 kg; E _k = 1/2 mv ² ; v = (2 E _k /m) ^1/2 ; $\text{v(Ag)}={\left(2\times \frac{44.33{\text{kg-m}}^2}{{\text{s}}^2}\times \frac{1}{2.056\text{kg}}\right)}^{1/2}=6.567=6.6\text{m/s}$

Note that calculation shows the speeds of the Al and Ag spheres at impact to be equal, as stated in Exercise 1.4(d).

Units and Measurement (Section 1.5)

1.31

1. 1 × 10 ⁻¹
2. 1 × 10 ⁻²
3. 1 × 10 ⁻¹⁵
4. 1 × 10 ⁻⁶
5. 1 × 10 ⁶
6. 1 × 10 ³
7. 1 × 10 ⁻⁹
8. 1 × 10 ⁻³
9. 1 × 10 ⁻¹²

1.32

1. $2.3\times {10}^{-10}\text{L}\times \frac{\text{1}\text{nL}}{1\times {10}^{-9}\text{L}}=0.23\text{nL}$
2. $4.7\times {10}^{-6}\text{g}\times \frac{1\mu \text{g}}{1\times {10}^{-6}\text{g}}=4.7\mu \text{g}$
3. $1.85\times {10}^{-12}\text{m}\times \frac{1\text{pm}}{1\times {10}^{-12}\text{m}}=1.85\text{pm}$
4. $16.7\times {10}^6\text{s}\times \frac{1\text{Ms}}{1\times {10}^6\text{s}}=16.7\text{Ms}$
5. $15.7\times {10}^3\text{g}\times \frac{1\text{kg}}{1\times {10}^3\text{g}}=15.7\text{kg}$
6. $1.34\times {10}^{-3}\text{m}\times \frac{1\text{mm}}{1\times {10}^{-3}\text{m}}=1.34\text{mm}$
7. $1.84\times {10}^2\text{cm}\times \frac{1\text{m}}{1\times {10}^2\text{cm}}=1.84\text{m}$

1.33

1. °C = 5/9 (°F – 32 °); 5/9 (72 – 32) = 22 °C
2. °F = 9/5 (°C) + 32 °; 9/5 (216.7) + 32 = 422.1 °F
3. K = °C + 273.15; 233 °C + 273.15 = 506 K
4. °C = 315 K – 273.15 = 41.85 = 42 °C; °F = 9/5 (41.85 °C) + 32 = 107 °F
5. °C = 5/9 (°F – 32); 5/9 (2500 – 32) = 1371 = 1400 °C

   K = 1371 °C + 273.15 = 1644 = 1600 K

6. °C = 0 K – 273.15 = –273.15 °C; °F = 9/5 (–273.15 °C) + 32 = –459.67 °F (assuming 0 K has infinite sig figs)

1.34

1. °C = 5/9 (87 °F – 32 °) = 31 °C
2. K = 25 °C + 273.15 = 298 K; °F = 9/5 (25 °C) + 32 = 77 °F
3. °C = 5/9 (400 °F – 32 °) = 204.444 = 200 °C

   K = °C + 273.15 = 204.444 °C + 273.15 = 477.59 = 500 K

4. °C = 77 K – 273.15 = –196.15 = –196 °C

   °F = 9/5 (–196.15 °C) + 32 = –321.07 = –321 °F

1.35

1. $\text{density}=\frac{\text{mass}}{\text{volume}}=\frac{40.55\text{g}}{25.0\text{mL}}=1.62\text{g/mL}\text{or}1.62{\text{g/cm}}^3$

   (The units cm ³ and mL will be used interchangeably in this manual.)

   Tetrachloroethylene, 1.62 g/mL, is more dense than water, 1.00 g/mL; tetrachloroethylene will sink rather than float on water.

2. $25.0{\text{cm}}^3\times 0.469\frac{\text{g}}{{\text{cm}}^{\text{3}}}=11.7\text{g}$

1.36

1. volume = length ³ (cm ³ ); density = mass/volume (g/cm ³ )

   volume = (1.500) ³ cm ³ = 3.375 cm ³

   $\text{density}=\frac{76.31\text{g}}{3.375{\text{cm}}^3}=22.61{\text{g/cm}}^{\text{3}}\text{osmium}$
2. $125.0\text{mL}\times \frac{1{\text{cm}}^3}{1\text{mL}}\times \frac{4.51\text{g}}{1{\text{cm}}^3}=563.75=564\text{g}\text{titanium}$
3. $0.1500\text{L}\times \frac{1\text{mL}}{1\times {10}^{-3}\text{L}}\times \frac{0.8787\text{g}}{1\text{mL}}=131.8\text{g}\text{benzene}$

1.37

1. $\text{calculated}\text{density}=\frac{38.5\text{g}}{45\text{mL}}=0.86\text{g/mL}$

   The substance is probably toluene, density = 0.866 g/mL.

2. $45.0\text{g}\times \frac{1\text{mL}}{1.114\text{g}}=40.4\text{mL}\text{ethylene}\text{glycol}$
3. A 100 mL graduated cylinder such as the one in Figure 1.21 usually has 1-mL markings. One can read the volume with certainty to the nearest mL, and estimate tenths of a mL. The volume calculated in has uncertainty in the tenths place, so a graduated cylinder like this will provide the appropriate accuracy of measurement.
4. ${(5.00)}^3{\text{cm}}^3\times \frac{8.90\text{g}}{1{\text{cm}}^3}=1.11\times {10}^3\text{g}(1.11\text{kg})\text{nickel}$

1.38

1. $\frac{\text{21}\text{.95}\text{g}}{\text{25}\text{.0}\text{mL}}\text{=}\text{0}\text{.878}\text{g/mL}$

   The tabulated value has four significant figures, while the experimental value has three. The tabulated value rounded to three figures is 0.879. The values agree within one in the last significant figure of the experimental value; the two results agree. The liquid could be benzene.

2. $15.0\text{g}\times \frac{1\text{mL}}{0.7781\text{g}}=19.3\text{mL}\text{cyclohexane}$

1. r = d/2 = 5.0 cm/2 = 2.5 cm

   V = 4/3 π r ³ = 4/3 × π × (2.5) ³ cm ³ = 65.4498 = 65 cm ³

   $65.4498{\text{cm}}^3\times \frac{11.34\text{g}}{{\text{cm}}^3}=7.4\times {10}^2\text{g}$

   (The answer has two significant figures because the diameter had only two significant figures.)

1.39 $36\text{billion}\text{metric}\text{tons}\times \frac{1\times {10}^9\text{metric}\text{tons}}{1\text{billion}\text{metric}\text{tons}}\times \frac{1000\text{kg}}{1\text{metric}\text{ton}}\times \frac{1000\text{g}}{1\text{kg}}=3.6\times {10}^{16}\text{g}$

The metric prefix for 1 × 10 ¹⁵ is peta, abbreviated P.

$3.6\times {10}^{16}\text{g}\times \frac{1\text{Pg}}{1\times {10}^{15}\text{g}}=36\text{Pg}$

1.40

1. The wafers have the same diameter as the boule, so the question becomes “how many 0.75 mm wafers can be cut from the 2 m boule?”

   $\frac{2.0\text{m}}{\text{boule}}\times \frac{1\text{mm}}{1\times {10}^{-3}\text{m}}\times \frac{1\text{wafer}}{0.75\text{mm}}=2667=2.7\times {10}^3\text{wafers}$

   As a practical matter, the thickness of the cutting blade reduces the actual number of disks that can be produced, so the real number is something less than 2667 wafers. Perhaps a more realistic answer, to 2 sig figs, is 2.6 × 10 ³ wafers.

2. Calculate the volume of the wafer in cm ³ . $\text{V}={\text{r}}^2\text{h}$

   $\text{r}=\frac{\text{d}}{2}=\frac{300\text{mm}}{2}\times \frac{1\text{cm}}{10\text{mm}}=15\text{cm};\text{h}=0.75\text{mm}\times \frac{1\text{cm}}{10\text{mm}}=7.5\times {10}^{-2}\text{cm}$

   $\text{V}=\pi {\text{r}}^2\text{h}=\pi {(15\text{cm})}^2(7.5\times {10}^{-2}\text{cm})=53.0144=53{\text{cm}}^3$

   Density = mass / V;    mass = density × V

   $\frac{2.33\text{g}}{{\text{cm}}^3}\times 53.0144{\text{cm}}^3=123.52=1.2\times {10}^2\text{g}$

1.41 Analyze .  Given: heat capacity of water = 1 Btu/lb-°F Find: J/Btu

Plan . $\text{heat capacity of water}=\frac{1\text{cal}}{\text{g-}{}^{\text{o}}\text{C}};\frac{\text{cal}}{\text{g-}{}^{\text{o}}\text{C}}\to \frac{\text{J}}{\text{g-}{}^{\text{o}}\text{C}}\to \frac{\text{J}}{\text{lb-}{}^{\text{o}}\text{F}}\to \frac{\text{J}}{\text{Btu}}$

This strategy requires changing °F to °C. Because this involves the magnitude of a degree on each scale, rather than a specific temperature, the 32 in the temperature relationship is not needed. 100 °C = 180 °F; 5 °C = 9 °F

Solve . $\frac{\text{1}\text{cal}}{\text{g-}{}^{\text{o}}\text{C}}\times \frac{\text{4}\text{.184}\text{J}}{\text{cal}}\times \frac{\text{453}\text{.6}\text{g}}{\text{lb}}\times \frac{\text{5}{}^{\text{o}}\text{C}}{\text{9}{}^{\text{o}}\text{F}}\times \frac{\text{1 lb-}{}^{\text{o}}\text{F}}{\text{1}\text{Btu}}=1054\text{J/Btu}$

1.42

1. Analyze .  Given: 1 kwh; 1 watt = 1 J/s; 1 watt-s = 1 J. Find: conversion factor for joules and kwh.

   Plan .  kwh → wh → ws → J

   Solve . $1\text{kwh}\times \frac{1000\text{w}}{1\text{kw}}\times \frac{60\text{min}}{\text{h}}\times \frac{60\text{s}}{\text{min}}\times \frac{1\text{J}}{1\text{w-s}}=3.6\times {10}^6\text{J}$

   1 kwh = 3.6 × 10 ⁶ J

2. Analyze .  Given: 100 watt bulb. Find: heat in kcal radiated by bulb or person in 24 hr.

   Plan .  1 watt = 1 J/s; 1 kcal = 4.184 × 10 ³ J; watt → J/s → J → kcal. Solve .

   $100\text{watt}=\frac{\text{100}\text{J}}{\text{1}\text{s}}\times \frac{60\text{sec}}{\text{min}}\times \frac{60\text{min}}{\text{hr}}\times 24\text{hr}\times \frac{\text{1}\text{kcal}}{\text{4}\text{.184}\times {\text{10}}^{\text{3}}\text{J}}=2065=2.1\times {10}^3\text{kcal}$

   24 hr has 2 sig figs, but 100 watt is ambiguous. The answer to 1 sig fig would be 2 × 10 ³ kcal.

Uncertainty in Measurement (Section 1.6)

1.43  Exact: (b), (d), and (f) (All others depend on measurements and standards that have margins of error, e.g., the length of a week as defined by the Earth’s rotation.)

1.44  Exact: (b), (e) (The number of students is exact on any given day.)

1.45

1. 3
2. 2
3. 5
4. 3
5. 5
6. 1 [See Sample Exercise 1.7 (c)]

1.46

1. 4
2. 3
3. 4
4. 5
5. 6
6. 2

1.47

1. 1.025 × 10 ²
2. 6.570 × 10 ²
3. 8.543 × 10 ⁻³
4. 2.579 × 10 ⁻⁴
5. –3.572 × 10 ⁻²

1.48

1. 7.93 × 10 ³ mi
2. 4.001 × 10 ⁴ km

1.49

1. 14.3505 + 2.65 = 17.0005 = 17.00 (For addition and subtraction, the minimum number of decimal places, here two, determines decimal places in the result.)
2. 952.7 – 140.7389 = 812.0
3. (3.29 × 10 ⁴ )(0.2501) = 8.23 × 10 ³ (For multiplication and division, the minimum number of significant figures, here three, determines sig figs in the result.)
4. 0.0588/0.677 = 8.69 × 10 ⁻²

1.50

1. [320.5 – 6104.5/2.3] = –2.3 × 10 ³ (The intermediate result has two significant figures, so only the thousand and hundred places in the answer are significant.)
2. [285.3 × 10 ⁵ – 0.01200 × 10 ⁵ ] × 2.8954 = 8.260 × 10 ⁷ (Because subtraction depends on decimal places, both numbers must have the same exponent to determine decimal places/sig figs. The intermediate result has 1 decimal place and 4 sig figs, so the answer has 4 sig figs.)

1. $\begin{array}{cccc}(0.0045\times \text{20,000}\text{.0})& +& (2813\times 12)& =3.4\times {10}^4\\ 2\text{sig figs /0 dec pl}& & 2\text{sig figs/first 2 digits}& \end{array}$

1. $\begin{array}{cccccc}\hfill 863& \times & [1255& -& (3.45\times 108)]& =7.60\times {10}^5\\ & & & & (3\text{sig figs /0 dec pl)}& \\ \hfill 3\text{sig figs}& \times & & \text{[0 dec pl/3 sig figs]}& & \text{= 3 sig figs}\end{array}$

1.51  The mass 21.427 g has 5 significant figures.

1.52  The volume in the graduated cylinder is 19.5 mL. Liquid volumes are read at the bottom of the meniscus, so the volume is slightly less than 20 mL. Volumes in this cylinder can be read with certainty to 1 mL, and with some uncertainty to 0.1 mL, so this measurement has 3 sig figs.

Dimensional Analysis (Section 1.7)

1.53  In each conversion factor, the old unit appears in the denominator, so it cancels, and the new unit appears in the numerator.

1. $\text{mm}\to \text{nm}:\frac{1\times {10}^{-3}\text{m}}{1\text{mm}}\times \frac{1\text{nm}}{1\times {10}^{-9}\text{m}}=1\times {10}^6\text{nm/mm}$
2. $\text{mg}\to \text{kg}:\frac{1\times {10}^{-3}\text{g}}{1\text{mg}}\times \frac{1\text{kg}}{1000\text{g}}=1\times {10}^{-6}\text{kg/mg}$
3. $\text{km}\to \text{ft}:\frac{1000\text{m}}{1\text{km}}\times \frac{1\text{cm}}{1\times {10}^{-2}\text{m}}\times \frac{1\text{in}}{2.54\text{cm}}\times \frac{1\text{ft}}{12\text{in}}$ = 3.28 × 10 ³ km/ft
4. ${\text{in}}^3\to {\text{cm}}^3:\frac{{(2.54)}^3{\text{cm}}^3}{1^3{\text{in}}^3}$ = 16.4 cm ³ /in ³

1.54  In each conversion factor, the old unit appears in the denominator, so it cancels, and the new unit appears in the numerator.

1. $\text{μm}\to \text{mm}\text{:}\frac{\text{1}\times {10}^{-6}\text{m}}{\text{1}\text{μm}}\times \frac{1\text{mm}}{\text{1}\times {10}^{-3}\text{m}}=\text{1}\times {10}^{-3}\text{mm/}\text{μm}$
2. $\text{ms}\to \text{ns}:\frac{1\times {10}^{-3}\text{s}}{1\text{ms}}\times \frac{1\text{ns}}{1\times {10}^{-9}\text{s}}=1\times {10}^6\text{ns/ms}$
3. mi → km: 1.6093 km/mi
4. ${\text{ft}}^3\to \text{L}:\frac{{(12)}^3{\text{in}}^3}{1{\text{ft}}^3}\times \frac{{(2.54)}^3{\text{cm}}^3}{1{\text{in}}^3}\times \frac{1\text{L}}{1000{\text{cm}}^3}=28.3\text{L/}{\text{ft}}^3$

1.55

1. $\frac{15.2\text{m}}{\text{s}}\times \frac{1\text{km}}{1000\text{m}}\times \frac{60\text{s}}{1\text{min}}\times \frac{60\text{min}}{1\text{hr}}=54.7\text{km/hr}$
2. $5.0\times {10}^3\text{L}\times \frac{1\text{gal}}{3.7854\text{L}}=1.3\times {10}^3\text{gal}$

1. $151\text{ft}\times \frac{1\text{yd}}{3\text{ft}}\times \frac{1\text{m}}{1.0936\text{yd}}=46.025=46.0\text{m}$

1. $\frac{60.0\text{cm}}{\text{d}}\times \frac{1\text{in}}{2.54\text{cm}}\times \frac{1\text{d}}{24\text{hr}}=0.984\text{in/hr}$

1.56

1. $\frac{2.998\times {10}^8\text{m}}{\text{s}}\times \frac{1\text{km}}{1000\text{m}}\times \frac{1\text{mi}}{1.6093\text{km}}\times \frac{60\text{s}}{1\text{min}}\times \frac{60\text{min}}{1\text{hr}}=6.707\times {10}^8\text{mi/hr}$
2. $1454\text{ft}\times \frac{1\text{yd}}{3\text{ft}}\times \frac{1\text{m}}{1.0936\text{yd}}=443.18=443.2\text{m}$
3. $3,666,500{\text{m}}^3\times \frac{1^3{\text{dm}}^3}{{(1\times {10}^{-1})}^3{\text{m}}^3}\times \frac{1\text{L}}{1{\text{dm}}^3}=3.6665\times {10}^9\text{L}$
4. $\frac{242\text{mg}\text{cholesterol}}{100\text{mL}\text{blood}}\times \frac{1\text{mL}}{1\times {10}^{-3}\text{L}}\times 5.2\text{L}\times \frac{1\times {10}^{-3}\text{g}}{1\text{mg}}=12.58=13\text{g}\text{cholesterol}$

1.57

1. $5.00\text{days}\times \frac{24\text{hr}}{1\text{day}}\times \frac{60\text{min}}{1\text{hr}}\times \frac{60\text{s}}{1\text{min}}=4.32\times {10}^5\text{s}$
2. $0.0550\text{mi}\times \frac{1.6093\text{km}}{\text{mi}}\times \frac{1000\text{m}}{1\text{km}}=88.5\text{m}$
3. $\frac{\$1.89}{\text{gal}}\times \frac{1\text{gal}}{3.7854\text{L}}=\frac{\$0.499}{\text{L}}$
4. $\frac{0.510\text{in}}{\text{ms}}\times \frac{2.54\text{cm}}{1\text{in}}\times \frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\times \frac{1\text{km}}{1000\text{m}}\times \frac{1\text{ms}}{1\times {10}^{-3}\text{s}}\times \frac{60\text{s}}{1\text{min}}\times \frac{60\text{min}}{1\text{hr}}=\frac{46.6\text{km}}{\text{hr}}$

   Estimate: 0.5 × 2.5 = 1.25; 1.25 × 0.01 ≈ 0.01; 0.01 × 60 × 60 ≈ 36 km/hr

5. $\frac{22.50\text{gal}}{\text{min}}\times \frac{3.7854\text{L}}{\text{gal}}\times \frac{1\text{min}}{60\text{s}}=1.41953=1.420\text{L/s}$

   Estimate: 20 × 4 = 80; 80/60 ≈ 1.3 L/s

6. $0.02500{\text{ft}}^3\times \frac{{12}^3{\text{in}}^3}{1{\text{ft}}^3}\times \frac{{2.54}^3{\text{cm}}^3}{1{\text{in}}^3}=707.9{\text{cm}}^3$

   Estimate: 10 ³ = 1000; 3 ³ = 27; 1000 × 27 = 27,000; 27,000/0.04 ≈ 700 cm ³

1.58

1. $0.105\text{in}\times \frac{2.54\text{cm}}{\text{in}}\times \frac{1\times {10}^{-2}\text{m}}{\text{cm}}\times \frac{1\text{mm}}{1\times {10}^{-3}\text{m}}=2.667=2.67\text{mm}$
2. $0.650\text{qt}\times \frac{1\text{L}}{1.057\text{qt}}\times \frac{1\text{mL}}{1\times {10}^{-3}\text{L}}=614.94=615\text{mL}$

1. $\frac{8.75\text{μm}}{\text{s}}\times \frac{1\times {10}^{-6}\text{m}}{1\text{μm}}\times \frac{1\text{km}}{1\times {10}^3\text{m}}\times \frac{60\text{s}}{1\text{min}}\times \frac{60\text{min}}{1\text{hr}}=3.15\times {10}^{-5}\text{km/hr}$

1. $1.955{\text{m}}^3\times \frac{{(1.0936)}^3{\text{yd}}^3}{1{\text{m}}^3}=2.55695=2.557{\text{yd}}^3$

1. $\frac{\$3.99}{\text{lb}}\times \frac{2.205\text{lb}}{1\text{kg}}=8.798=\$8.80\text{/kg}$

1. $\frac{8.75\text{lb}}{{\text{ft}}^3}\times \frac{453.59\text{g}}{1\text{lb}}\times \frac{1{\text{ft}}^3}{{12}^3{\text{in}}^3}\times \frac{1{\text{in}}^3}{{2.54}^3{\text{cm}}^3}\times \frac{1{\text{cm}}^3}{1\text{mL}}=0.140\text{g/mL}$

1.59

1. $31\text{gal}\times \frac{4\text{qt}}{1\text{gal}}\times \frac{1\text{L}}{1.057\text{qt}}=1.2\times {10}^2\text{L}$

   Estimate: (30 × 4)/1 ≈ 120 L

2. $\frac{6\text{mg}}{\text{kg}\text{(body)}}\times \frac{1\text{kg}}{2.205\text{lb}}\times 185\text{lb}=5\times {10}^2\text{mg}$

   Estimate: 6/2 = 3; 3 × 180 = 540 mg

3. $\frac{400\text{km}}{47.3\text{L}}\times \frac{1\text{mi}}{1.6093\text{km}}\times \frac{1\text{L}}{1.057\text{qt}}\times \frac{4\text{qt}}{1\text{gal}}=\frac{19.9\text{mi}}{\text{gal}}$

   (2 × 10 ¹ mi/gal for 1 sig fig)

   Estimate: 400/50 = 8; 8/1.6 = 5; 5/1 = 5; 5 × 4 ≈ 20 mi/gal

4. $200\text{cups}\times \frac{1\text{lb}}{50\text{cups}}\times \frac{1\text{kg}}{2.205\text{lb}}=1.81\text{kg}$

   (2 kg for 1 sig fig)

   Estimate: 1 lb = 50 cups, 4 lb = 200 cups; 4 lb ≈ 2 kg

1.60

1. $1257\text{m}\text{i}\times \frac{1\text{k}\text{m}}{0.62137\text{m}\text{i}}\times \frac{\text{c}\text{h}\text{a}\text{r}\text{g}\text{e}}{225\text{k}\text{m}}=8.99\text{c}\text{h}\text{a}\text{r}\text{g}\text{e}\text{s}$

   Because charges are integral events, 9 total charges are required. The trip begins with a full charge, so 8 additional charges during the trip are needed.

2. $\frac{14\text{m}}{\text{s}}\times \frac{1\text{km}}{1\times {10}^3\text{m}}\times \frac{1\text{mi}}{1.6093\text{km}}\times \frac{60\text{s}}{1\text{min}}\times \frac{60\text{min}}{1\text{hr}}=31\text{mi/hr}$
3. $450{\text{in}}^3\times \frac{{(2.54)}^3{\text{cm}}^3}{1{\text{in}}^3}\times \frac{1\text{mL}}{1{\text{cm}}^3}\times \frac{1\times {10}^{-3}\text{L}}{1\text{mL}}=7.37\text{L}$
4. $2.4\times {10}^5\text{barrels}\times \frac{42\text{gal}}{1\text{barrel}}\times \frac{4\text{qt}}{1\text{gal}}\times \frac{1\text{L}}{1.057\text{qt}}=3.8\times {10}^7\text{L}$

1.61  14.5 ft × 16.5 ft × 8.0 ft = 1914 = 1.9 × 10 ³ ft ³ (2 sig figs)

$1914{\text{ft}}^3\times \frac{{(1\text{yd})}^3}{{(3\text{ft})}^3}\times \frac{{(1\text{m})}^3}{{(1.0936)}^3{\text{yd}}^3}\times \frac{{10}^3{\text{dm}}^3}{1{\text{m}}^3}\times \frac{1\text{L}}{1{\text{dm}}^3}\times \frac{1.19\text{g}}{\text{L}}\times \frac{1\text{kg}}{1000\text{g}}=64.4985=64\text{kg}\text{air}$

Estimate: 1900/27 ≈ 60; (60 × 1)/1 ≈ 60 kg

1.62  10.6 ft × 14.8 ft × 20.5 ft = 3216.04 = 3.22 × 10 ³ ft ³

$3216.04{\text{ft}}^3\times \frac{{(1\text{yd})}^3}{{(3\text{ft})}^3}\times \frac{{(1\text{m})}^3}{{(1.0936\text{yd})}^3}\times \frac{48\text{μg}\text{Co}}{1{\text{m}}^3}\times \frac{1\times {10}^{-6}\text{g}}{1\text{μg}}=4.4\times {10}^{-3}\text{g}\text{Co}$

1.63  Strategy:

1. Calculate volume of gold (Au) in cm ³ in the sheet
2. Mass = density × volume
3. Change g $\to$ troy oz and $

$100\text{ft}\times 82\text{ft}\times \frac{{(12)}^2{\text{in}}^2}{1{\text{ft}}^2}\times 5\times {10}^{-6}\text{in}\times \frac{{(2.54)}^3{\text{cm}}^3}{1{\text{in}}^3}=96.75=1\times {10}^2{\text{cm}}^3\text{Au}$

$96.75{\text{cm}}^3\text{Au}\times \frac{19.32\text{g}}{1{\text{cm}}^3}\times \frac{1\text{troy}\text{oz}}{31.1034768\text{g}}\times \frac{\$1654}{\text{troy}\text{oz}}=\$99,399=\$1\times {10}^5$

(Strictly speaking, the datum 100 ft has 1 sig fig, so the result has 1 sig fig.)

1.64  A wire is a very long, thin cylinder of volume, V = π r ² h, where h is the length of the wire and π r ² is the cross-sectional area of the wire.

Strategy:

1. Calculate total volume of copper in cm ³ from mass and density
2. $\text{h (length in cm)}=\frac{\text{V}}{\pi {\text{r}}^2}$
3. Change cm → ft

$150\text{lb}\text{Cu}\times \frac{453.6\text{g}}{1\text{lb}\text{Cu}}\times \frac{1{\text{cm}}^3}{8.94\text{g}}=7610.7=7.61\times {10}^3{\text{cm}}^3$

$\text{r}=\text{d/}2=7.50\text{mm}\times \frac{1\text{cm}}{10\text{mm}}\times \frac{1}{2}=0.375\text{cm}$

$\text{h}=\frac{\text{V}}{\pi {\text{r}}^2}=\frac{7610.7{\text{cm}}^3}{\pi {(0.375)}^2{\text{cm}}^2}=1.7227\times {10}^4=1.72\times {10}^4\text{cm}$

$1.7227\times {10}^4\text{cm}\times \frac{1\text{in}}{2.54\text{cm}}\times \frac{1\text{ft}}{12\text{in}}=565\text{ft}$

(too difficult to estimate)

Additional Exercises

1.65

1. A gold ingot is pure metallic gold, a pure substance.
2. A cup of coffee is a solution if there are no suspended solids (coffee grounds). It is a heterogeneous mixture if there are grounds. If cream or sugar is added, the homogeneity of the mixture depends on how thoroughly the components are mixed.
3. A wood plank is a heterogeneous mixture of various cellulose components. The different domains in the mixture are visible as wood grain or knots.

1.66

1. A hypothesis is more likely to eventually be shown to be incorrect. A hypothesis is a possible explanation based on preliminary experimental data. A theory may be more general, and has a significant body of experimental evidence to support it.
2. A law reliably predicts the behavior of matter, while a theory provides an explanation for that behavior.

1.67  According to the law of constant composition, any sample of vitamin C has the same relative amount of carbon and oxygen; the ratio of oxygen to carbon in the isolated sample is the same as the ratio in synthesized vitamin C.

$\frac{\text{2}\text{.00}\text{g}\text{O}}{\text{1}\text{.50}\text{g}\text{C}}=\frac{\text{x}\text{g}\text{O}}{\text{6}\text{.35}\text{g}\text{C}};\text{x}=\frac{\text{(2}\text{.00}\text{g}\text{O)}\text{(6}\text{.35}\text{g}\text{C)}}{\text{1}\text{.50}\text{g}\text{C}}=8.47\text{g}\text{O}$

1.68

1. (l) → (g)
2. °F = 9/5 (°C) + 32 °; 9/5 (12) + 32 = 53.6 = 54 °F
3. 0.765 g/mL × 103.5 mL = 79.178 = 79.2 g ethyl chloride

1.69

1. 1. (22.52 + 22.48 + 22.54)/3 = 22.51
   2. (22.64 + 22.58 + 22.62)/3 = 22.61

   Based on the average, set I is more accurate. That is, it is closer to the true value of 22.52%.

2. Average $\text{deviation}=\text{Σ}\left|\text{v}\text{a}\text{l}\text{u}\text{e}\text{-}\text{a}\text{v}\text{e}\text{r}\text{a}\text{g}\text{e}\right|\text{/}3$
   1. | 22.52 – 22.51 | + |22.48 – 22.51 | + |22.54 – 22.51 |/3 = 0.02
   2. | 22.64 – 22.61 | + |22.58 – 22.61 | + |22.62 – 22.61 |/3 = 0.02

   Based on average deviations, the two sets display the same precision, even though set I is more accurate. [According to Section 1.5, standard deviation is a measure that is often used to determine precision. Using the formula for calculating standard deviation given in Appendix A.5, the values for the two sets are 0.03 and 0.03, respectively. The standard deviations of the two sets are also the same, confirming that the two sets are equally precise.]

1.70

1. Inappropriate. The circulation number indicates a precision of one part per million, 1 ppm. First, the term “circulation” is not specific, and can include paid subscriptions, news stand sales and other means of distribution. Even if we consider paid subscriptions only, it is unlikely that the number of paid subscriptions is know to a precision of 1 ppm.

1. Inappropriate. The population number implies a precision of 1 part per million, 1 ppm. Census data are taken only once every ten years and cannot directly measure every citizen even in the year the census is taken. Population fluctuates daily, so this precision is not reasonable.

1. Appropriate. The percentage has three significant figures. In a population as large as that of the United States, the number of people named Brown can surely be counted by census data or otherwise to a precision of three significant figures.

1. Inappropriate. Letter grades are posted at most to two decimal places and three significant figures (if plus and minus modifiers are quantified). The grade-point-average, obtained by addition and division, cannot have more decimal places or significant figures than the numbers being averaged.

1.71

1. volume
2. area
3. volume
4. density
5. time
6. length
7. temperature

1.72

1. $\frac{\text{m}}{{\text{s}}^2}$
2. $\frac{\text{k}\text{g}\text{-}\text{m}}{{\text{s}}^2}$
3. $\frac{\text{k}\text{g}\text{-}\text{m}}{{\text{s}}^2}\times \text{m}=\frac{\text{k}\text{g}\text{-}{\text{m}}^2}{{\text{s}}^2}$
4. $\frac{\text{kg-m}}{{\text{s}}^{\text{2}}}\times \frac{\text{1}}{{\text{m}}^{\text{2}}}=\frac{\text{kg}}{{\text{m-s}}^{\text{2}}}$
5. $\frac{{\text{kg-m}}^2}{{\text{s}}^2}\times \frac{1}{\text{s}}=\frac{{\text{kg-m}}^2}{{\text{s}}^3}$
6. $\frac{\text{m}}{\text{s}}$
7. $\text{kg}\times {\left(\frac{\text{m}}{\text{s}}\right)}^2=\frac{{\text{kg-m}}^{\text{2}}}{{\text{s}}^{\text{2}}}$

1.73

1. $2.4\times {10}^5\text{mi}\times \frac{\text{1}\text{.609}\text{km}}{\text{1}\text{mi}}\times \frac{\text{1000}\text{m}}{\text{1}\text{km}}=3.862\times {10}^8=3.9\times {10}^8\text{m}$
2. $2.4\times {10}^5\text{mi}\times \frac{1.609\text{km}}{1\text{mi}}\times \frac{\text{1}\text{hr}}{\text{350}\text{km}}\times \frac{\text{60}\text{min}}{\text{1}\text{hr}}\times \frac{60\text{s}}{1\text{min}}=4.0\times {10}^6\text{s}$
3. $3.862\times {10}^8\text{m}\times 2\times \frac{1\text{s}}{3.00\times {10}^8\text{m}}=2.574=2.6\text{s}$
4. $\frac{\text{29}\text{.783}\text{km}}{\text{s}}\times \frac{\text{1}\text{mi}}{\text{1}\text{.6093}\text{km}}\times \frac{\text{60}\text{s}}{\text{1}\text{min}}\times \frac{\text{60}\text{min}}{\text{1}\text{hr}}=6.6624\times {10}^4\text{mi/hr}$

1.74

1. Baking powder is not a pure substance. It is a mixture of basic and acidic compounds that, in the presence of water, react to form CO ₂ which causes baked goods to rise.
2. Lemon juice is not a pure substance. It is a mixture of water, citric acid, and other natural flavors. Its exact composition depends on the characteristics of the lemon that produces the juice.
3. Propane is a nearly pure substance. Propane itself is odorless, but tank propane contains trace amounts of an odor-producing substance so that leaks can easily be detected.
4. Aluminum foil is a nearly pure substance.
5. Ibuprofen itself is a pure substance. Ibuprofen tablets probably contain the active ingredient and other binders, drying agents and coatings.

1. Bourbon whiskey is not a pure substance. It is a mixture of water, alcohol, and flavoring compounds.

1. Helium gas is a pure substance.

1. Clear water pumped from a deep aquifer is a nearly pure substance. All natural water (not distilled or deionized) contains trace minerals.

1.75

1. $575\text{ft}\times \frac{\text{12}\text{in}}{\text{1}\text{ft}}\times \frac{\text{2}\text{.54}\text{cm}}{\text{1}\text{in}}\times \frac{\text{10}\text{mm}}{\text{1}\text{cm}}\times \frac{\text{1}\text{quarter}}{\text{1}\text{.55}\text{mm}}=1.1307\times {10}^5=1.13\times {10}^5\text{quarters}$
2. $1.1307\times {10}^5\text{quarters}\times \frac{5.67\text{g}}{1\text{quarter}}=6.41\times {10}^5\text{g}\text{(641}\text{kg)}$
3. $1.1307\times {10}^5\text{quarters}\times \frac{\text{1}\text{dollar}}{\text{4}\text{quarters}}=\text{\$}28,268=\text{\$}2.83\times {10}^4$
4. $\text{\$}16,213,166,914,811.11\times \frac{\text{1}\text{stack}}{\text{\$28,268}}=5.7355\times {10}^8=5.74\times {10}^8\text{stacks}$

1.76

1. $\frac{\text{\$1950}}{\text{acre-ft}}\times \frac{\text{1}\text{acre}}{\text{4840}{\text{yd}}^{\text{2}}}\times \frac{\text{3}\text{ft}}{\text{1}\text{yd}}\times \frac{{\text{(1}\text{.094}\text{yd)}}^{\text{3}}}{{\text{(1}\text{m)}}^{\text{3}}}\times \frac{{\text{(1}\text{m)}}^{\text{3}}}{{\text{(10}\text{dm)}}^{\text{3}}}\times \frac{{\text{(1}\text{dm)}}^{\text{3}}}{\text{1}\text{L}}=\$1.583\times {10}^{-3}/\text{L or}0.1583¢\text{/L (0}\text{.158}¢\text{/L to}3\text{sig figs)}$
2. $\frac{\text{\$1950}}{\text{acre-ft}}\times \frac{\text{1}\text{acre-ft}}{\text{2}\text{households-year}}\times \frac{\text{1}\text{year}}{\text{365}\text{days}}\times 1\text{household}\text{=}\frac{\text{\$2}\text{.671}}{\text{day}}\text{=}\frac{\text{\$2}\text{.67}}{\text{day}}$

1.77  Select a common unit for comparison, in this case the kg.

1 kg > 2 lb, 1 L ≈ 1 qt

5 lb potatoes < 2.5 kg

5 kg sugar = 5 kg

1 gal = 4 qt ≈ 4 L; 1 mL H ₂ O = 1 g H ₂ O; 1 L = 1000 g, 4 L = 4000 g = 4 kg

The order of mass from lightest to heaviest is 5 lb potatoes < 1 gal water < 5 kg sugar.

1.78  There are 347 degrees between the freezing and boiling points on the oleic acid (O) scale and 100 degrees on the celsius (C) scale. Also, 13 °C = 0 °O. By analogy with °F and °C,

${}^{\text{o}}\text{O}=\frac{100}{347}({}^{\text{o}}\text{C}-13)\text{or}{}^{\text{o}}\text{C}=\frac{347}{100}({}^{\text{o}}\text{O})+13$

These equations correctly relate the freezing point (and boiling point) of oleic acid on the two scales.

$\text{f}\text{.p}\text{.}\text{of}{\text{H}}_{\text{2}}\text{O:}{}^{\text{o}}\text{O}=\frac{100}{347}(0{}^{\text{o}}\text{C}-13)=-3.746=-4{}^{\text{o}}\text{O}$

1.79  The most dense liquid, Hg, will sink; the least dense, cyclohexane, will float; H ₂ O will be in the middle.

1.80  Density is the ratio of mass and volume. For samples with the same volume, in this case spheres with the same diameter, the denser ball will have a greater mass. The heavier ball, the red one on the right in the diagram is more dense.

1.81  The mass of water in the bottle does not change with temperature, but the density (ratio of mass to volume) does. That is, the amount of volume occupied by a certain mass of water changes with temperature. Calculate the mass of water in the bottle at 25 °C, and then the volume occupied by this mass at −10 °C.

1. $25{}^{\text{o}}\text{C}:1.50\text{L}{\text{H}}_{\text{2}}\text{O}\times \frac{\text{1000}{\text{cm}}^{\text{3}}}{\text{1}\text{L}}\times \frac{\text{0}\text{.997}\text{g}{\text{H}}_{\text{2}}\text{O}}{\text{1}{\text{cm}}^{\text{3}}}$ = 1.4955 × 10 ³ = 1.50 × 10 ³ g H ₂ O

   −10 °C: 1.4955 × 10 ³ g H ₂ O × $\frac{1\text{c}{\text{m}}^3}{0.917\text{g}{\text{H}}_2\text{O}}\times \frac{1\text{L}}{1000\text{c}{\text{m}}^3}$ = 1.6309 = 1.63 L

2. No. If the soft-drink bottle is completely filled with 1.50 L of water, the 1.63 L of ice cannot be contained in the bottle. The extra volume of ice will push through any opening in the bottle, or crack the bottle to create an opening.

1.82  mass of toluene = 58.58 g – 32.65 g = 25.93 g

$\text{volume}\text{of}\text{toluene}=25.93\text{g}\times \frac{\text{1}\text{mL}}{\text{0}\text{.864}\text{g}}=30.0116=30.0\text{mL}$

volume of solid = 50.00 mL – 30.0116 mL = 19.9884 = 20.0 mL

$\text{density}\text{of}\text{solid}=\frac{\text{32}\text{.65}\text{g}}{\text{19}\text{.9884}\text{mL}}=1.63\text{g/mL}$

1.83  V = 4/3 π r ³ = 4/3 π (28.9 cm) ³ = 1.0111 × 10 ⁵ = 1.01 × 10 ⁵ = 1.01 × 10 ⁵ cm ³

$1.0111\times {10}^5{\text{cm}}^3\times \frac{\text{19}\text{.3}\text{g}}{{\text{cm}}^{\text{3}}}\times \frac{\text{1}\text{lb}}{\text{453}\text{.59}\text{g}}=4302=4.30\times {10}^3\text{lb}$

No, the thief is unlikely to be able to carry the sphere without assistance. It weighs 4300 pounds, more than two tons.

1.84 $1.00\text{gal}\text{battery}\text{acid}\times \frac{\text{4}\text{qt}}{\text{1}\text{gal}}\times \frac{\text{1000}\text{mL}}{\text{1}\text{.0567}\text{qt}}\times \frac{\text{1}\text{.28}\text{g}}{\text{mL}}=4845.3=4.85\times {10}^3\text{g}\text{battery}\text{acid}$

$4.8453\times {10}^3\text{g}\text{battery}\text{acid}\times \frac{\text{38}\text{.1}\text{g}\text{sulfuric}\text{acid}}{\text{100}\text{g}\text{battery}\text{acid}}=1846=1.85\times {10}^3\text{g}\text{sulfuric}\text{acid}$

1.85

1. $\frac{40\text{lb}\text{peat}}{14\times 20\times 30{\text{in}}^3}\times \frac{1{\text{in}}^3}{{(2.54)}^3{\text{cm}}^3}\times \frac{453.6\text{g}}{1\text{lb}}=0.13{\text{g/cm}}^{\text{3}}\text{peat}$

   $\frac{\text{40}\text{lb}\text{soil}}{\text{1}\text{.9}\text{gal}}\times \frac{\text{1}\text{gal}}{\text{4}\text{qt}}\times \frac{\text{1}\text{.057}\text{qt}}{\text{1}\text{L}}\times \frac{1\times {10}^{-3}\text{L}}{1\text{mL}}\times \frac{1\text{mL}}{1{\text{cm}}^3}\times \frac{453.6\text{g}}{1\text{lb}}=2.5{\text{g/cm}}^3\text{soil}$

   No. Volume must be specified to compare mass. The densities tell us that a certain volume of peat moss is “lighter” (weighs less) than the same volume of top soil.

2. 1 bag peat = 14 × 20 × 30 = 8.4 × 10 ³ in ³

   $15.0\text{ft}\times 20.0\text{ft}\times 3.0\text{in}\times \frac{{12}^2{\text{in}}^2}{{\text{ft}}^2}=129,600=1.3\times {10}^5{\text{in}}^3\text{peat}\text{needed}$

   $129,600{\text{in}}^3\times \frac{1\text{bag}}{8.4\times {10}^3{\text{in}}^3}=15.4=15\text{bags}\text{(Buy}\text{16}\text{bags}\text{of}\text{peat}\text{.)}$

1.86 $8.0\text{oz}\times \frac{1\text{lb}}{16\text{oz}}\times \frac{453.6\text{g}}{\text{lb}}\times \frac{1{\text{cm}}^3}{2.70\text{g}}=84.00=84{\text{cm}}^3$

$\frac{\text{84}{\text{cm}}^{\text{3}}}{\text{50}{\text{ft}}^{\text{2}}}\times \frac{{\text{1}}^{\text{2}}{\text{ft}}^{\text{2}}}{{\text{12}}^{\text{2}}{\text{in}}^{\text{2}}}\times \frac{{\text{1}}^{\text{2}}{\text{in}}^{\text{2}}}{\text{2}{\text{.54}}^{\text{2}}{\text{cm}}^{\text{2}}}\times \frac{\text{10}\text{mm}}{\text{1}\text{cm}}=0.018\text{mm}$

1.87  The total solar flux, the power provided by the sun, is the average solar flux times the area of the disc in sunlight at any instant. However, because the earth rotates, each disc is in sunlight for only half of the day, so only half of this flux can be collected.

$\frac{\text{680}\text{W}}{{\text{m}}^{\text{2}}}\times 1.28\times {10}^{14}{\text{m}}^2\times \text{1/}2=4.352\times {10}^{16}=4.4\times {10}^{16}\text{W}$

Collection is 10% efficient, so only 10% of this power is available, $4.4\times {10}^{15}\text{W}$

$15\text{TW}\times \frac{1\times {10}^{12}\text{W}}{1\text{TW}}=15\times {10}^{12}\text{W}\text{needed}$

$\frac{15\times {10}^{12}\text{W}\text{needed}}{4.352\times {10}^{15}\text{W}\text{available}}\times 100=0.3447=0.34\%$ Earth’s surface needed

1.88  Warren’s first hypothesis was that the bacterium he observed was involved in causing ulcers. This was a radical hypothesis in 1979, when it was commonly accepted that the acidic environment of the stomach would not support microorganisms. Strong evidence was needed to change the opinion and practice of the medical community.

To test his hypothesis, Warren first collected many more tissue samples, from patients with and without ulcers. He established that healthy tissue did not contain the bacterium, while many of the ulcerated samples did. From there, with Marshall and other collaborators, he established two different ways to diagnose the infection (determine the presence of the bacterium in patients with potential ulcers), and learned to culture (grow in the laboratory) the new bacterium so that its effects could be studied. Finally, in patients diagnosed and treated for bacterial ulcers, he demonstrated that the recurrence of ulcers was rare.

This information was obtained from the autobiography of J. Robin Warren, http://www.nobelprize.org/nobel_prizes/medicine/laureates/2005/warren-autobio.php

1.89 $45.23\text{g}\text{ethanol}\times \frac{1{\text{cm}}^3}{0.789\text{g}\text{ethanol}}=57.3257=57.33{\text{cm}}^3,\text{volume}\text{of}\text{cylinder}$

$\begin{array}{l}\text{V}={\text{πr}}^2\text{h};\text{r}={\left(\text{V}/\text{πh}\right)}^{1/2}=\left[\frac{57.3257{\text{cm}}^3}{\text{π}\times 25.0\text{cm}}\right]=0.854338=0.854\text{cm}\\ \text{d}=2\text{r}=1.71\text{cm}\end{array}$

1.90

1. Let x = mass of Au in jewelry

   9.85 – x = mass of Ag in jewelry

   The total volume of jewelry = volume of Au + volume of Ag

   $0.675{\text{cm}}^3=\text{x}\text{g}\times \frac{1{\text{cm}}^3}{19.3\text{g}}+(9.85-\text{x})\text{g}\times \frac{1{\text{cm}}^3}{10.5\text{g}}$

   $0.675=\frac{\text{x}}{\text{19}\text{.3}}+\frac{9.85-\text{x}}{10.5}\text{(To solve, multiply both sides by}(19.3)(10.5))$

   $\begin{array}{lll}\hfill 0.675\left(19.3\right)\left(10.5\right)& =\hfill & 10.5\text{x}+\left(9.85-\text{x}\right)\left(19.3\right)\hfill \\ \hfill 136.79& =\hfill & 10.5\text{x}+190.105-19.3\text{x}\hfill \\ \hfill -53.315& =\hfill & -8.8\text{x}\hfill \end{array}$

   x = 6.06 g Au; 9.85 g total – 6.06 g Au = 3.79 g Ag

   $\text{mass}\%\text{Au}=\frac{6.06\text{g}\text{Au}}{9.85\text{g}\text{jewelry}}\times 100=61.5\%\text{Au}$

2. 24 carats × 0.615 = 15 carat gold

1.91  The separation with distinctly separated red and blue spots is more successful. The procedure that produced the purple blur did not separate the two dyes. To quantify the characteristics of the separation, calculate a reference value for each spot that is

$$\frac{\text{distance}\text{traveled}\text{by}\text{spot}}{\text{distance}\text{traveled}\text{by}\text{solvent}}$$

If the values for the two spots are fairly different, the separation is successful. (One could measure the distance between the spots, but this would depend on the length of paper used and be different for each experiment. The values suggested above are independent of the length of paper.)

1.92

1. False. Air and water are not elements. Air is a homogeneous mixture of gases and pure water is a compound.
2. False. Mixtures can contain any number of pure substances, either elements or compounds or both.
3. True.
4. True.
5. False. When yellow stains in a kitchen sink are treated with bleach water, a chemical change occurs.
6. True.
7. False. The number 0.0033 has the same number of significant figures as 0.033.
8. True. (In a conversion factor, the quantity in the numerator is equal to the quantity in the denominator, so the overall numerical value is one.)
9. True.

1.93  The densities are:

carbon tetrachloride (methane, tetrachloro) – 1.5940 g/cm ³

hexane – 0.6603 g/cm ³

benzene – 0.87654 g/cm ³

methylene iodide (methane, diiodo) – 3.3254 g/cm ³

Only methylene iodide will separate the two granular solids. The undesirable solid (2.04 g/cm ³ ) is less dense than methylene iodide and will float; the desired material is more dense than methylene iodide and will sink. The other three liquids are less dense than both solids and will not produce separation.

1.94

1. $10.0\text{mg}\times \frac{1\times {10}^{-3}\text{g}}{1\text{mg}}\times \frac{1{\text{cm}}^3}{0.20\text{g}}=0.050{\text{cm}}^3=0.050\text{mL}\text{volume}$
2. $10.0\text{mg}\times \frac{1\times {10}^{-3}\text{g}}{1\text{mg}}\times \frac{1242{\text{m}}^2}{1\text{g}}=12.42=12.4{\text{m}}^2\text{surface}\text{area}$
3. 7.748 mg Hg initial – 0.001 mg Hg remain = 7.747 mg Hg removed

   $\frac{7.747\text{mg}\text{Hg}\text{removed}}{7.748\text{mg}\text{Hg}\text{initial}}\times 100=99.99\%\text{Hg}\text{removed}$

4. 10.0 mg “spongy” initial + 7.747 mg Hg removed = 17.747 = 17.7 mg after exposure

 

2 Atoms, Molecules, and Ions

Visualizing Concepts

2.1

1. Like charges repel and opposite charges attract, so the sign of the electrical charge on the particle is negative.
2. The greater the magnitude of the charges, the greater the electrostatic repulsion or attraction. As the charge on the plates is increased, the bending will increase.
3. As the mass of the particle increases and speed stays the same, linear momentum (mv) of the particle increases and bending decreases. (See A Closer Look : The Mass Spectrometer.)

2.2

1. $\text{\%}\text{abundance}=\frac{\#\text{of mass number}\times \text{particles}}{\text{total number of particles}}\times 100$

   12 red ²⁹³ Nv particles

   8 blue ²⁹⁵ Nv particles

   20 total particles

   $\text{\% abundance}{}^{293}\text{Nv}=\frac{12}{20}\times 100=60\text{\%}$

   $\text{\% abundance}{}^{295}\text{Nv}=\frac{8}{20}\times 100=40\text{\%}$

2. Atomic weight (AW) is the same as average atomic mass.

   Atomic weight (average atomic mass) = Σ fractional abundance mass of isotope

   AW of Nv = 0.60(293.15) + 0.40(295.15) = 293.95 amu

   (Because % abundance was calculated by counting exact numbers of particles, assume % abundance is an exact number. Then, the number of significant figures in the AW is determined by the number of sig figs in the masses of the isotopes.)

2.3   In general, metals occupy the left side of the chart, and nonmetals the right side.

metals : red and green nonmetals : blue and yellow

alkaline earth metal : red noble gas : yellow

2.4   Because the number of electrons (negatively charged particles) does not equal the number of protons (positively charged particles), the particle is an ion. The charge on the ion is 2–.

Atomic number = number of protons = 16. The element is S, sulfur.

Mass number = protons + neutrons = 32

${}_{16}^{32}{\text{S}}^{2-}$

2.5   In a solid, particles are close together and their relative positions are fixed. In a liquid, particles are close but moving relative to each other. In a gas, particles are far apart and moving. Most ionic compounds are solids because of the strong forces among charged particles. Molecular compounds can exist in any state: solid, liquid, or gas.

Because the molecules in ii are far apart, ii must be a molecular compound. The particles in i are near each other and exist in a regular, ordered arrangement, so i is likely to be an ionic compound.

2.6   Formula: IF ₅ Name: iodine pentafluoride

Because the compound is composed of elements that are all nonmetals, it is molecular.

2.7   See Figure 2.17. yellow box: 1+ (group 1A); blue box: 2+ (group 2A)

black box: 3+ (a metal in Group 3A); red box: 2– (a nonmetal in group 6A);

green box: 1– (a nonmetal in group 7A)

2.8   Cations (red spheres) have positive charges; anions (blue spheres) have negative charges. There are twice as many anions as cations, so the formula has the general form CA ₂ . Only Ca(NO ₃ ) ₂ , calcium nitrate, is consistent with the diagram.

2.9   These two compounds are isomers. They have the same chemical formula, C ₄ H ₉ Cl, but different arrangements of atoms. That is, they have different chemical structures. In the first isomer, the Cl atom is bound to the second C atom from the left. In the second isomer, the Cl atom is bound to the right-most C atom.

2.10

1. In the absence of an electric field, there is no electrostatic interaction between the oil drops and the apparatus, so the rate of fall of the oil drops is determined solely by the force of gravity. In the presence of an electric field, there is electrostatic attraction between the negatively charged oil drops and the positively charged plate, as well as electrostatic repulsion between the negatively charged oil drops and the negative plate. These electrostatic forces oppose the force of gravity and change the rate of fall of the drops.
2. Each individual drop has a different number of electrons associated with it. The greater the accumulated negative charge on the drop, the greater the electrostatic forces between the oil drop and the plates. If the combined electrostatic forces are greater than the force of gravity, the drop moves up.

The Atomic Theory of Matter and the Discovery of Atomic Structure (Sections 2.1 and 2.2)

2.11

1. $\text{ratio of masses}=\frac{0.727\text{g O}}{0.273\text{g C}}=2.663=2.66$
2. $\text{ratio of masses}=\frac{0.571\text{g O}}{0.429\text{g C}}=1.331=1.33$
3. The two mass ratios are related by a factor of 2. In the first compound, CO ₂ , twice as much O is bound to one gram of C as in the second compound. The empirical formula of the second compound is then CO.

2.12

1. 6.500 g compound – 0.384 g hydrogen = 6.116 g sulfur
2. Conservation of mass

2.13

1. $\frac{\text{17}\text{.60 g oxygen}}{\text{30}\text{.82 g nitrogen}}=\frac{\text{0}\text{.5711 g O}}{\text{1 g N}};\text{0}\text{.5711/0}\text{.5711}=\text{1}\text{.0}$

   $\frac{\text{35}\text{.20 g oxygen}}{\text{30}\text{.82 g nitrogen}}=\frac{\text{1}\text{.142 g O}}{\text{1 g N}};\text{1}\text{.142/0}\text{.5711}=\text{2}\text{.0}$

   $\frac{\text{70}\text{.40 g oxygen}}{\text{30}\text{.82 g nitrogen}}=\frac{\text{2}\text{.284 g O}}{\text{1 g N}};\text{2}\text{.284/0}\text{.5711}=\text{4}\text{.0}$

   $\frac{\text{88}\text{.00 g oxygen}}{\text{30}\text{.82 g nitrogen}}=\frac{\text{2}\text{.855 g O}}{\text{1 g N}};\text{2}\text{.855/0}\text{.5711}=\text{5}\text{.0}$

2. These masses of oxygen per one gram nitrogen are in the ratio of 1:2:4:5 and thus obey the law of multiple proportions . Multiple proportions arise because atoms are the indivisible entities combining, as stated in Dalton’s theory. Because atoms are indivisible, they must combine in ratios of small whole numbers.

2.14

1. 1: $\frac{\text{3}\text{.56 g fluorine}}{\text{4}\text{.75 g iodine}}=\text{0}\text{.749 g fluorine/1 g iodine}$

   2: $\frac{\text{3}\text{.43 g fluorine}}{\text{7}\text{.64 g iodine}}=\text{0}\text{.449 g fluorine/1 g iodine}$

   3: $\frac{\text{9}\text{.86 g fluorine}}{\text{9}\text{.41 g iodine}}=\text{1}\text{.05 g fluorine/1 g iodine}$

2. To look for integer relationships among these values, divide each one by the smallest. If the quotients aren’t all integers, multiply by a common factor to obtain all integers.

   1: 0.749/0.449 = 1.67; 1.67 × 3 = 5

   2: 0.449/0.449 = 1.00; 1.00 × 3 = 3

   3: 1.05/0.449 = 2.34; 2.34 × 3 = 7

   The ratio of g fluorine to g iodine in the three compounds is 5:3:7. These are in the ratio of small whole numbers and, therefore, obey the law of multiple proportions . This integer ratio indicates that the combining fluorine “units” (atoms) are indivisible entities.